### 数学代写|凸优化作业代写Convex Optimization代考|Examples of convex sets

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## 数学代写|凸优化作业代写Convex Optimization代考|Hyperplanes and halfspaces

A hyperplane is an affine set (hence a convex set) and is of the form
$$H=\left{\mathbf{x} \mid \mathbf{a}^{T} \mathbf{x}=b\right} \subset \mathbb{R}^{n},$$
where $\mathbf{a} \in \mathbb{R}^{n} \backslash\left{\mathbf{0}_{n}\right}$ is a normal vector of the hyperplane, and $b \in \mathbb{R}$. Analytically it is the solution set of a linear equation of the components of $\mathbf{x}$. In geometrical sense, a hyperplane can be interpreted as the set of points having a constant inner product (b) with the normal vector (a). Since affdim $(H)=n-1$,

the hyperplane (2.30) can also be expressed as
$$H=\operatorname{aff}\left{\mathbf{s}{1}, \ldots, \mathbf{s}{n}\right} \subset \mathbb{R}^{n},$$
where $\left{\mathbf{s}{1}, \ldots, \mathbf{s}{n}\right} \subset H$ is any affinely independent set. Then it can be seen that
$$\left{\begin{array}{l} \mathbf{B} \triangleq\left[\mathbf{s}{2}-\mathbf{s}{1}, \ldots, \mathbf{s}{n}-\mathbf{s}{1}\right] \in \mathbb{R}^{n \times(n-1)} \text { and } \operatorname{dim}(\mathcal{R}(\mathbf{B}))=n-1 \ \mathbf{B}^{T} \mathbf{a}=\mathbf{0}{n-1}, \text { i.e. }, \mathcal{R}(\mathbf{B})^{\perp}=\mathcal{R}(\mathbf{a}) \end{array}\right.$$ implying that the normal vector a can be determined from $\left{\mathbf{s}{1}, \ldots, \mathbf{s}_{n}\right}$ up to a scale factor.

The hyperplane $H$ defined in $(2.30)$ divides $\mathbb{R}^{n}$ into two closed halfspaces as follows:
\begin{aligned} &H_{-}=\left{\mathbf{x} \mid \mathbf{a}^{T} \mathbf{x} \leq b\right} \ &H_{+}=\left{\mathbf{x} \mid \mathbf{a}^{T} \mathbf{x} \geq b\right} \end{aligned}
and so each of them is the solution set of one (non-trivial) linear inequality. Note that $\mathbf{a}=\nabla\left(\mathbf{a}^{T} \mathbf{x}\right)$ denotes the maximally increasing direction of the linear function $\mathbf{a}^{T} \mathbf{x}$. The above representations for both $H_{-}$and $H_{+}$for a given $\mathbf{a} \neq \mathbf{0}$, are not unique, while they are unique if $\mathbf{a}$ is normalized such that $|\mathbf{a}|_{2}=1$. Moreover, $H_{-} \cap H_{+}=H$.
An open halfspace is a set of the form
$$H_{–}=\left{\mathbf{x} \mid \mathbf{a}^{T} \mathbf{x}b\right}$$
where $\mathbf{a} \in \mathbb{R}^{n}, \mathbf{a} \neq \mathbf{0}$, and $b \in \mathbb{R}$.

## 数学代写|凸优化作业代写Convex Optimization代考|Euclidean balls and ellipsoids

A Euclidean ball (or, simply, ball) in $\mathbb{R}^{n}$ has the following form:
$$B\left(\mathbf{x}{c}, r\right)=\left{\mathbf{x} \mid\left|\mathbf{x}-\mathbf{x}{c}\right|_{2} \leq r\right}=\left{\mathbf{x} \mid\left(\mathbf{x}-\mathbf{x}{c}\right)^{T}\left(\mathbf{x}-\mathbf{x}{c}\right) \leq r^{2}\right},$$
where $r>0$. The vector $\mathbf{x}{c}$ is the center of the ball and the positive scalar $r$ is its radius (see Figure 2.7). The Euclidean ball is also a 2-norm ball, and, for simplicity, a ball without explicitly mentioning the associated norm, means the Euclidean ball hereafter. Another common representation for the Euclidean ball is $$B\left(\mathbf{x}{c}, r\right)=\left{\mathbf{x}{c}+r \mathbf{u} \mid|\mathbf{u}|{2} \leq 1\right}$$
It can be easily proved that the Euclidean ball is a convex set.
Proof of convexity: Let $\mathbf{x}{1}$ and $\mathbf{x}{2} \in B\left(\mathbf{x}{c}, r\right)$, i.e., $\left|\mathbf{x}{1}-\mathbf{x}{c}\right|{2} \leq r$ and $| \mathbf{x}{2}-$ $\mathbf{x}{c} |_{2} \leq r$. Then,
\begin{aligned} \left|\theta \mathbf{x}{1}+(1-\theta) \mathbf{x}{2}-\mathbf{x}{c}\right|{2} &=\left|\theta \mathbf{x}{1}+(1-\theta) \mathbf{x}{2}-\left[\theta \mathbf{x}{c}+(1-\theta) \mathbf{x}{c}\right]\right|_{2} \ &=\left|\theta\left(\mathbf{x}{1}-\mathbf{x}{c}\right)+(1-\theta)\left(\mathbf{x}{2}-\mathbf{x}{c}\right)\right|_{2} \ & \leq\left|\theta\left(\mathbf{x}{1}-\mathbf{x}{c}\right)\right|_{2}+\left|(1-\theta)\left(\mathbf{x}{2}-\mathbf{x}{c}\right)\right|_{2} \ & \leq \theta r+(1-\theta) r \ &=r, \text { for all } 0 \leq \theta \leq 1 \end{aligned}
Hence, $\theta \mathbf{x}{1}+(1-\theta) \mathbf{x}{2} \in B\left(\mathbf{x}{c}, r\right)$ for all $\theta \in[0,1]$, and thus we have proven that $B\left(\mathbf{x}{c}, r\right)$ is convex.

A related family of convex sets are ellipsoids (see Figure $2.7$ ), which have the form
$$\mathcal{E}=\left{\mathbf{x} \mid\left(\mathbf{x}-\mathbf{x}{c}\right)^{T} \mathbf{P}^{-1}\left(\mathbf{x}-\mathbf{x}{c}\right) \leq 1\right},$$
where $\mathbf{P} \in \mathcal{S}{++}^{n}$ and the vector $\mathbf{x}{c}$ is the center of the ellipsoid. The matrix $\mathbf{P}$ determines how far the ellipsoid extends in every direction from the center; the lengths of the semiaxes of $\mathcal{E}$ are given by $\sqrt{\lambda_{i}}$, where $\lambda_{i}$ are eigenvalues of $\mathbf{P}$. Note that a ball is an ellipsoid with $\mathbf{P}=r^{2} \mathbf{I}{n}$ where $r>0$. Another common representation of an ellipsoid is $$\mathcal{E}=\left{\mathbf{x}{c}+\mathbf{A u} \mid|\mathbf{u}|_{2} \leq 1\right}$$
where $\mathbf{A}$ is a square matrix and nonsingular. The ellipsoid $\mathcal{E}$ expressed by (2.38) is actually the image of the 2 -norm ball $B(\mathbf{0}, 1)=\left{\mathbf{u} \in \mathbb{R}^{n} \mid|\mathbf{u}|_{2} \leq 1\right}$ via an

affine mapping $\mathbf{x}{c}+\mathbf{A u}$ (cf. $\left.(2.58)\right)$, where $\mathbf{A}=\left(\mathbf{P}^{1 / 2}\right)^{T}$, and the proof will be presented below. With the expression (2.38) for an ellipsoid, the singular values of $\mathbf{A}, \sigma{i}(\mathbf{A})=\sqrt{\lambda_{i}}$ (lengths of semiaxes) characterize the structure of the ellipsoid in a more straightforward fashion than the expression (2.37).
Proof of the ellipsoid representation (2.38) and convexity: Let
$$\mathbf{P}=\mathbf{Q} \boldsymbol{\Lambda} \mathbf{Q}^{T}=\left(\mathbf{P}^{1 / 2}\right)^{T} \mathbf{P}^{1 / 2}$$
$(\mathrm{EVD}$ of $\mathbf{P} \succ \mathbf{0})$, where $\boldsymbol{\Lambda}=\boldsymbol{\operatorname { D i a g }}\left(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}\right)$ and
$$\mathbf{P}^{1 / 2}=\boldsymbol{\Lambda}^{1 / 2} \mathbf{Q}^{T},$$
in which $\Lambda^{1 / 2}=\operatorname{Diag}\left(\sqrt{\lambda_{1}}, \sqrt{\lambda_{2}}, \ldots, \sqrt{\lambda_{n}}\right)$. Then
$$\mathbf{P}^{-1}=\mathbf{Q} \boldsymbol{\Lambda}^{-1} \mathbf{Q}^{T}=\mathbf{P}^{-1 / 2}\left(\mathbf{P}^{-1 / 2}\right)^{T}, \quad \mathbf{P}^{-1 / 2}=\mathbf{Q} \mathbf{\Lambda}^{-1 / 2}$$
From the definition of ellipsoid, we have
\begin{aligned} \mathcal{E} &=\left{\mathbf{x} \mid\left(\mathbf{x}-\mathbf{x}{c}\right)^{T} \mathbf{P}^{-1}\left(\mathbf{x}-\mathbf{x}{c}\right) \leq 1\right} \ &=\left{\mathbf{x} \mid\left(\mathbf{x}-\mathbf{x}{c}\right)^{T} \mathbf{Q} \mathbf{\Lambda}^{-1} \mathbf{Q}^{T}\left(\mathbf{x}-\mathbf{x}{c}\right) \leq 1\right} \end{aligned}
Let $\mathbf{z}=\mathbf{x}-\mathbf{x}{c}$. Then $$\mathcal{E}=\left{\mathbf{x}{c}+\mathbf{z} \mid \mathbf{z}^{T} \mathbf{Q} \mathbf{\Lambda}^{-1} \mathbf{Q}^{T} \mathbf{z} \leq 1\right}$$
Now, by letting $\mathbf{u}=\mathbf{\Lambda}^{-1 / 2} \mathbf{Q}^{T} \mathbf{z}=\left(\mathbf{P}^{-1 / 2}\right)^{T} \mathbf{z}$, we then obtain
$$\mathcal{E}=\left{\mathbf{x}{c}+\left(\mathbf{P}^{1 / 2}\right)^{T} \mathbf{u} \mid|\mathbf{u}|{2} \leq 1\right}$$
which is exactly (2.38) with $\left(\mathbf{P}^{1 / 2}\right)^{T}$ replaced by $\mathbf{A}$.

## 数学代写|凸优化作业代写Convex Optimization代考|Polyhedra

A polyhedron is a nonempty convex set and is defined as the solution set of a finite number of linear equalities and inequalities:
\begin{aligned} \mathcal{P} &=\left{\mathbf{x} \mid \mathbf{a}{i}^{T} \mathbf{x} \leq b{i}, i=1,2, \ldots, m, \mathbf{c}{j}^{T} \mathbf{x}=d{j}, j=1,2, \ldots, p\right} \ &=\left{\mathbf{x} \mid \mathbf{A} \mathbf{x} \preceq \mathbf{b}=\left(b_{1}, \ldots, b_{m}\right), \mathbf{C x}=\mathbf{d}=\left(d_{1}, \ldots, d_{p}\right)\right} \end{aligned}
where ” $\preceq “$ stands for componentwise inequality, $\mathbf{A} \in \mathbb{R}^{m \times n}$ and $\mathbf{C} \in \mathbb{R}^{p \times n}$ are matrices whose rows are $\mathbf{a}{j}^{T} \mathrm{~s}$ and $\mathbf{c}{j}^{T} \mathrm{~s}$, respectively, and $\mathbf{A} \neq \mathbf{0}$ or $\mathbf{C} \neq \mathbf{0}$ must be true. Note that either $m=0$ or $p=0$ is allowed as long as the other parameter is finite and nonzero.

A polyhedron is just the intersection of some halfspaces and hyperplanes (see Figure 2.8). A polyhedron can be unbounded, while a bounded polyhedron is called a polytope, e.g., any 1-norm ball and $\infty$-norm ball of finite radius are polytopes.

Remark 2.3 It can be seen that the $\ell$-dimensional space $\mathbb{R}^{\ell}$ (also an affine set) cannot be expressed in the standard form (2.41) with either nonzero $\mathbf{A}$ or nonzero $\mathbf{C}$, and so it is not a polyhedron for any $\ell \in \mathbb{Z}{++}$. However, because any affine set in $\mathbb{R}^{\ell}$ can be expressed as (2.7) and each $\mathbf{x}$ in the set satisfies (2.8), this implies that the affine set must be a polyhedron if its affine dimension is strictly less than $\ell$. For instance, any subspaces with dimension less than $n$ in $\mathbb{R}^{n}$, and any hyperplane that is defined by a normal vector $\mathbf{a} \neq \mathbf{0}$ and a point $\mathbf{x}{0}$ on the hyperplane, i.e.,
$$\mathcal{H}\left(\mathbf{a}, \mathbf{x}{0}\right)=\left{\mathbf{x} \mid \mathbf{a}^{T}\left(\mathbf{x}-\mathbf{x}{0}\right)=0\right}=\left{\mathbf{x}_{0}\right}+\mathcal{R}(\mathbf{a})^{\perp}$$
in $\mathbb{R}^{n}$, and rays, line segments, and halfspaces are all polyhedra.

## 数学代写|凸优化作业代写Convex Optimization代考|Hyperplanes and halfspaces

H=\left{\mathbf{x} \mid \mathbf{a}^{T} \mathbf{x}=b\right} \subset \mathbb{R}^{n}，H=\left{\mathbf{x} \mid \mathbf{a}^{T} \mathbf{x}=b\right} \subset \mathbb{R}^{n}，

H=\operatorname{aff}\left{\mathbf{s}{1}, \ldots, \mathbf{s}{n}\right} \subset \mathbb{R}^{n},H=\operatorname{aff}\left{\mathbf{s}{1}, \ldots, \mathbf{s}{n}\right} \subset \mathbb{R}^{n},

$$\left{乙≜[s2−s1,…,sn−s1]∈Rn×(n−1) 和 暗淡⁡(R(乙))=n−1 乙吨一种=0n−1, IE ,R(乙)⊥=R(一种)\对。$$ 暗示法线向量 a 可以从\left{\mathbf{s}{1}, \ldots, \mathbf{s}_{n}\right}\left{\mathbf{s}{1}, \ldots, \mathbf{s}_{n}\right}高达一个比例因子。

\begin{aligned} &H_{-}=\left{\mathbf{x} \mid \mathbf{a}^{T} \mathbf{x} \leq b\right} \ &H_{+}=\left{\ mathbf{x} \mid \mathbf{a}^{T} \mathbf{x} \geq b\right} \end{对齐}\begin{aligned} &H_{-}=\left{\mathbf{x} \mid \mathbf{a}^{T} \mathbf{x} \leq b\right} \ &H_{+}=\left{\ mathbf{x} \mid \mathbf{a}^{T} \mathbf{x} \geq b\right} \end{对齐}

H_{–}=\left{\mathbf{x} \mid \mathbf{a}^{T} \mathbf{x}b\right}H_{–}=\left{\mathbf{x} \mid \mathbf{a}^{T} \mathbf{x}b\right}

## 数学代写|凸优化作业代写Convex Optimization代考|Euclidean balls and ellipsoids

B\left(\mathbf{x}{c}, r\right)=\left{\mathbf{x} \mid\left|\mathbf{x}-\mathbf{x}{c}\right|_{ 2} \leq r\right}=\left{\mathbf{x} \mid\left(\mathbf{x}-\mathbf{x}{c}\right)^{T}\left(\mathbf{x }-\mathbf{x}{c}\right) \leq r^{2}\right},B\left(\mathbf{x}{c}, r\right)=\left{\mathbf{x} \mid\left|\mathbf{x}-\mathbf{x}{c}\right|_{ 2} \leq r\right}=\left{\mathbf{x} \mid\left(\mathbf{x}-\mathbf{x}{c}\right)^{T}\left(\mathbf{x }-\mathbf{x}{c}\right) \leq r^{2}\right},

|θX1+(1−θ)X2−XC|2=|θX1+(1−θ)X2−[θXC+(1−θ)XC]|2 =|θ(X1−XC)+(1−θ)(X2−XC)|2 ≤|θ(X1−XC)|2+|(1−θ)(X2−XC)|2 ≤θr+(1−θ)r =r, 对全部 0≤θ≤1

\mathcal{E}=\left{\mathbf{x} \mid\left(\mathbf{x}-\mathbf{x}{c}\right)^{T} \mathbf{P}^{-1} \left(\mathbf{x}-\mathbf{x}{c}\right) \leq 1\right},\mathcal{E}=\left{\mathbf{x} \mid\left(\mathbf{x}-\mathbf{x}{c}\right)^{T} \mathbf{P}^{-1} \left(\mathbf{x}-\mathbf{x}{c}\right) \leq 1\right},

(和在D的磷≻0)， 在哪里Λ=诊断(λ1,λ2,…,λn)和

\begin{对齐} \mathcal{E} &=\left{\mathbf{x} \mid\left(\mathbf{x}-\mathbf{x}{c}\right)^{T} \mathbf{P }^{-1}\left(\mathbf{x}-\mathbf{x}{c}\right) \leq 1\right} \ &=\left{\mathbf{x} \mid\left(\mathbf {x}-\mathbf{x}{c}\right)^{T} \mathbf{Q} \mathbf{\Lambda}^{-1} \mathbf{Q}^{T}\left(\mathbf{ x}-\mathbf{x}{c}\right) \leq 1\right} \end{对齐}\begin{对齐} \mathcal{E} &=\left{\mathbf{x} \mid\left(\mathbf{x}-\mathbf{x}{c}\right)^{T} \mathbf{P }^{-1}\left(\mathbf{x}-\mathbf{x}{c}\right) \leq 1\right} \ &=\left{\mathbf{x} \mid\left(\mathbf {x}-\mathbf{x}{c}\right)^{T} \mathbf{Q} \mathbf{\Lambda}^{-1} \mathbf{Q}^{T}\left(\mathbf{ x}-\mathbf{x}{c}\right) \leq 1\right} \end{对齐}

\mathcal{E}=\left{\mathbf{x}{c}+\left(\mathbf{P}^{1 / 2}\right)^{T} \mathbf{u} \mid|\mathbf{ u}|{2} \leq 1\right}\mathcal{E}=\left{\mathbf{x}{c}+\left(\mathbf{P}^{1 / 2}\right)^{T} \mathbf{u} \mid|\mathbf{ u}|{2} \leq 1\right}

## 数学代写|凸优化作业代写Convex Optimization代考|Polyhedra

\begin{aligned} \mathcal{P} &=\left{\mathbf{x} \mid \mathbf{a}{i}^{T} \mathbf{x} \leq b{i}, i=1, 2, \ldots, m, \mathbf{c}{j}^{T} \mathbf{x}=d{j}, j=1,2, \ldots, p\right} \ &=\left{\ mathbf{x} \mid \mathbf{A} \mathbf{x} \preceq \mathbf{b}=\left(b_{1}, \ldots, b_{m}\right), \mathbf{C x}= \mathbf{d}=\left(d_{1}, \ldots, d_{p}\right)\right} \end{对齐}\begin{aligned} \mathcal{P} &=\left{\mathbf{x} \mid \mathbf{a}{i}^{T} \mathbf{x} \leq b{i}, i=1, 2, \ldots, m, \mathbf{c}{j}^{T} \mathbf{x}=d{j}, j=1,2, \ldots, p\right} \ &=\left{\ mathbf{x} \mid \mathbf{A} \mathbf{x} \preceq \mathbf{b}=\left(b_{1}, \ldots, b_{m}\right), \mathbf{C x}= \mathbf{d}=\left(d_{1}, \ldots, d_{p}\right)\right} \end{对齐}

\mathcal{H}\left(\mathbf{a}, \mathbf{x}{0}\right)=\left{\mathbf{x} \mid \mathbf{a}^{T}\left(\mathbf {x}-\mathbf{x}{0}\right)=0\right}=\left{\mathbf{x}_{0}\right}+\mathcal{R}(\mathbf{a})^ {\perp}\mathcal{H}\left(\mathbf{a}, \mathbf{x}{0}\right)=\left{\mathbf{x} \mid \mathbf{a}^{T}\left(\mathbf {x}-\mathbf{x}{0}\right)=0\right}=\left{\mathbf{x}_{0}\right}+\mathcal{R}(\mathbf{a})^ {\perp}

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